Question

During dehydration of alcohol to alkene by heating with conc. H\(_2\)SO\(_4\), the initiation step is:

• A. Formation of an ester
• B. Formation of carbocation
• C. Protonation of alcohol
• D. Elimination of water

Hint:

In acid-catalysed dehydration, the alcohol is first protonated to \(\mathrm{ROH_2^+}\), then water leaves to give a carbocation, and finally a proton is lost to give the alkene.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
When an alcohol is heated with concentrated \(\mathrm{H_2SO_4}\) at a high temperature (about 443 K for ethanol), it loses water and forms an alkene. This is acid-catalysed dehydration, and it runs in three steps. The question asks which step comes first.

Step 2: Write the mechanism for ethanol.
Step (a): the oxygen of the alcohol has lone pairs, so it picks up a proton from the acid.
\[ \mathrm{CH_3CH_2-\ddot{O}H + H^+ \rightleftharpoons CH_3CH_2-\overset{+}{O}H_2} \]
Step (b): the protonated alcohol loses a water molecule. A carbocation forms, and this is the slow step.
\[ \mathrm{CH_3CH_2-\overset{+}{O}H_2 \longrightarrow CH_3\overset{+}{C}H_2 + H_2O} \]
Step (c): the carbocation loses a proton from the neighbouring carbon, and the alkene forms. The acid is regenerated.
\[ \mathrm{CH_3\overset{+}{C}H_2 \longrightarrow CH_2{=}CH_2 + H^+} \]

Step 3: Pick the first step.
The first (initiation) step is the protonation of the alcohol, which is option (C).

Step 4: Why the other options are wrong.
Option (A): ester (alkyl hydrogen sulphate) formation is a side reaction at lower temperature, not a step of this mechanism.
Option (B): the carbocation forms only after the protonated alcohol loses water, so it is the second step.
Option (D): water leaves in step (b), after protonation. The \(\mathrm{-OH}\) group is a poor leaving group until it is protonated.

Final Answer:
The initiation step is the protonation of the alcohol. \[ \boxed{\text{Option 3 (C): Protonation of alcohol}} \]